Every shortest path from A to B must take exactly 3 steps to the right and 2 steps up — 5 steps in all, in some order. A path is completely decided the moment you choose which 2 of those 5 steps are the upward ones. The number of ways to choose 2 positions out of 5 is C(5, 2) = (5 · 4) / (2 · 1) = 10. Why this counts every path exactly once: each distinct choice of the two “up” steps gives a different route, and every route corresponds to one such choice. Takeaway: shortest grid paths taking r rights and u ups number C(r + u, u) — the same numbers that fill Pascal’s triangle.

Do not trust a first impression — walk the boundary yourself, and a quiet elegance appears: cutting the corner changes the perimeter not at all. Trace the edge of the L: two full outer sides of length 8, then the two short edges of the notch, each 4, and the two remaining outer pieces, each also 4. Add them in order around the shape: 8 + 8 + 4 + 4 + 4 + 4 = 32. Why it equals the whole square’s perimeter (4 · 8 = 32): the cut removes 4 + 4 of outer edge but restores exactly 4 + 4 of inner edge — a perfect exchange. Takeaway: for a rectangular notch cut into a corner, the perimeter is unchanged; let the careful path, not the eye, be your guide.

Motion yields to measurement — track time and distance step by step. First, how long does Ana run? She covers 4 miles at 8 mph, taking 4 ÷ 8 = 0.5 hour = 30 minutes, then stops. In those same 30 minutes Ben, at 6 mph, has covered 6 · 0.5 = 3 miles — so when Ana stops, Ben is still 4 − 3 = 1 mile behind her. Ben now needs to cover that last mile at 6 mph: 1 ÷ 6 hour = 1/6 · 60 = 10 minutes. Takeaway: in motion problems, convert everything to a common measure of time, find the remaining gap, then divide the gap by the speed that closes it.

A shape, you will find, always hides an arithmetic. Look at what the tilted square leaves behind: four identical right triangles in the corners, each with two short legs of 5 (half of a side of 10). Each corner triangle has area ½ · 5 · 5 = 12.5, and four of them together cover 4 · 12.5 = 50. The whole square measures 10 · 10 = 100, so the shaded square must be 100 − 50 = 50. The beautiful truth: joining the midpoints of any square yields an inner square with exactly half the area. (See it through my theorem as well: the tilted side is √(5² + 5²) = √50, so its square is 50.) When a figure resists you, count what surrounds it. The path around a problem is often shorter than the path through it — remember that, and geometry will open to you.

Two equations, two unknowns — eliminate one and the rest follows. We have x + 2y = 10 and 3x − y = 9. Double the second equation so the y terms can cancel with the first: 6x − 2y = 18. Add it to x + 2y = 10, and the y’s vanish: 7x = 28, so x = 4. Put x = 4 back into the first equation: 4 + 2y = 10, giving y = 3. Therefore x − y = 4 − 3 = 1. Takeaway: to solve a pair of linear equations, scale one until a variable’s coefficients match, then add or subtract to make that variable disappear.

You could never multiply 7 by itself two thousand times — and you never need to, because the last digit does not care about the rest of the number. It marches in a short, faithful cycle. Watch only the final digit as the powers grow: 7 → 9 (from 49) → 3 (from 343) → 1 (from 2401) → and then 7 again. The pattern 7, 9, 3, 1 repeats every four steps, forever. So I ask a single question: where in the cycle does the 2025th power fall? Divide the exponent by the cycle length: 2025 = 4 · 506 + 1, a remainder of 1. A remainder of 1 lands on the first member of the cycle — which is 7. The lesson: when a calculation looks impossibly large, hunt for the hidden rhythm. Finding the pattern where others see only size — that is the whole art of number theory, and it is within your reach.

You need not find the two sides at all — one identity carries you straight to the diagonal. Let the sides be l and w. The perimeter gives l + w = 34 ÷ 2 = 17, and the area gives l · w = 60. The diagonal, by the Pythagorean theorem, is √(l² + w²). Now use the identity (l + w)² = l² + 2lw + w²: so l² + w² = (l + w)² − 2lw = 17² − 2 · 60 = 289 − 120 = 169. The diagonal is √169 = 13. Takeaway: when you know a sum and a product, the sum of squares comes free from (sum)² − 2·(product) — no need to solve for the pieces.

Call the numbers a and b. Combining the reciprocals over a common denominator gives 1/a + 1/b = (a + b)/(ab). Since a + b = 15 and the reciprocal sum is 5/6, this reads 15/(ab) = 5/6, so ab = 15·6/5 = 18. Now the square-of-a-sum identity (a + b)² = a² + 2ab + b² rearranges to a² + b² = (a + b)² − 2ab = 15² − 2·18 = 225 − 36 = 189. Notice the numbers themselves are (15 ± 3√17)/2 — irrational — so trying to guess a whole-number pair fails; the symmetric-function route is the only clean one. Stopping at (a + b)² = 225 comes from the classic error of treating the square of a sum as the sum of the squares. Subtracting ab only once, 225 − 18 = 207, forgets the factor 2 in the identity, while a sign slip that adds 2ab gives 225 + 36 = 261. Flipping the combined fraction — writing ab/(a + b) = 5/6, so ab = 12.5 — leads to 225 − 25 = 200.

Translate the story into congruences: the total n satisfies n ≡ 5 (mod 8) and n ≡ 2 (mod 9). To find one number that works, list numbers leaving remainder 2 when divided by 9 — 2, 11, 20, 29 — and test each against the first condition: 29 = 3·8 + 5, so 29 works. Because 8 and 9 share no common factor, the Chinese Remainder Theorem says the solutions repeat every lcm(8, 9) = 72, so every valid total has the form 29 + 72k. Listing the ones strictly between 100 and 500 gives 101, 173, 245, 317, 389, 461 — six totals. A common slip is computing (461 − 101)/72 = 5 and forgetting to add 1 for the first term, an off-by-one that undercounts the list. Ignoring the lower bound and keeping 29 in the list gives one extra value, and ignoring both bounds so that 29 and 533 are both kept gives two extra. Finally, a student who assumes the pattern repeats every 36 instead of every 72 — using half the true period — counts twice as many totals as actually exist.